Difference between revisions of "2002 AMC 12B Problems/Problem 17"
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| + | {{duplicate|[[2002 AMC 12B Problems|2002 AMC 12B #17]] and [[2002 AMC 10B Problems|2002 AMC 10B #21]]}} | ||
== Problem == | == Problem == | ||
Andy’s lawn has twice as much area as Beth’s lawn and three times as much area as Carlos’ lawn. Carlos’ lawn mower cuts half as fast as Beth’s mower and one third as fast as Andy’s mower. If they all start to mow their lawns at the same time, who will finish first? | Andy’s lawn has twice as much area as Beth’s lawn and three times as much area as Carlos’ lawn. Carlos’ lawn mower cuts half as fast as Beth’s mower and one third as fast as Andy’s mower. If they all start to mow their lawns at the same time, who will finish first? | ||
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<math>\mathrm{(A)}\ \text{Andy} | <math>\mathrm{(A)}\ \text{Andy} | ||
\qquad\mathrm{(B)}\ \text{Beth} | \qquad\mathrm{(B)}\ \text{Beth} | ||
| − | \qquad\mathrm{(C)}\ \text{Carlos} | + | \qquad\mathrm{(C)}\ \text{Carlos} |
| − | + | \qquad\mathrm{(D)}\ \text{Andy\ and \ Carlos\ tie\ for\ first.} | |
\qquad\mathrm{(E)}\ \text{All\ three\ tie.}</math> | \qquad\mathrm{(E)}\ \text{All\ three\ tie.}</math> | ||
== Solution == | == Solution == | ||
| − | We say Andy's lawn has an area of <math>x</math>. Beth's lawn thus has an area of <math>\frac{x}2</math>, and Carlos's lawn has an area of <math>\frac{x}3</math>. | + | We say Andy's lawn has an area of <math>x</math>. Beth's lawn thus has an area of <math>\frac{x}{2}</math>, and Carlos's lawn has an area of <math>\frac{x}{3}</math>. |
| − | We say Andy's lawn mower cuts at a speed of <math>y</math>. Carlos's cuts at a speed of <math>\frac{y}3</math>, and Beth's cuts at a speed <math>\frac{2y}3</math>. | + | We say Andy's lawn mower cuts at a speed of <math>y</math>. Carlos's cuts at a speed of <math>\frac{y}{3}</math>, and Beth's cuts at a speed <math>\frac{2y}{3}</math>. |
| − | Each person's lawn is cut at a speed of <math>\frac{\text{area}}{\text{rate}}</math>, so Andy's is cut in <math>\frac{x}y</math> time, as is Carlos's. Beth's is cut in <math>\ | + | Each person's lawn is cut at a speed of <math>\frac{\text{area}}{\text{rate}}</math>, so Andy's is cut in <math>\frac{x}{y}</math> time, as is Carlos's. Beth's is cut in <math>\frac{3}{4}\times\frac{x}{y}</math>, so the first one to finish is <math>\boxed{\mathrm{(B)}\ \text{Beth}}</math>. |
== See also == | == See also == | ||
| + | {{AMC10 box|year=2002|ab=B|num-b=20|num-a=22}} | ||
{{AMC12 box|year=2002|ab=B|num-b=16|num-a=18}} | {{AMC12 box|year=2002|ab=B|num-b=16|num-a=18}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
Revision as of 15:23, 29 July 2011
- The following problem is from both the 2002 AMC 12B #17 and 2002 AMC 10B #21, so both problems redirect to this page.
Problem
Andy’s lawn has twice as much area as Beth’s lawn and three times as much area as Carlos’ lawn. Carlos’ lawn mower cuts half as fast as Beth’s mower and one third as fast as Andy’s mower. If they all start to mow their lawns at the same time, who will finish first?
Solution
We say Andy's lawn has an area of 
. Beth's lawn thus has an area of 
, and Carlos's lawn has an area of 
.
We say Andy's lawn mower cuts at a speed of 
. Carlos's cuts at a speed of 
, and Beth's cuts at a speed 
.
Each person's lawn is cut at a speed of 
, so Andy's is cut in 
time, as is Carlos's. Beth's is cut in 
, so the first one to finish is 
.
See also
| 2002 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 20 |
Followed by Problem 22 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
| 2002 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 16 |
Followed by Problem 18 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
